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May 13, 2014 at 17:14 comment added Mick @Blue I forgot that your diagram is different than mine. If I have the roles of B and C interchanged, then I need not ask the question. Thanx.
May 13, 2014 at 6:17 comment added Blue @Mick: $B^\prime$ is on $\overleftrightarrow{QS}$, which is the perpendicular bisector of $\overline{AC}$. Thus, $B^\prime$ is equidistant from $A$ and $C$, making $\triangle AB^\prime C$ isosceles; since there's a $60^\circ$ angle at $A$, that triangle is in fact equilateral. (Likewise, $C^\prime$ is on the perpendicular bisector of $\overline{AB}$, etc, etc.)
May 13, 2014 at 6:00 comment added Mick @Blue Please explain why, through the introduction of B’ (and also C’), you can infer that △AB′C is also equilateral. I understand that S is its ortho-center and S lies on the angle bisector of ∠A.
May 12, 2014 at 19:40 comment added Blue @rah4927: I was playing with a dynamic GeoGebra sketch of the problem, and it occurred to me that $R$ and $S$ were on the bisector of $\angle A$. Having constructed the bisector to verify my suspicion, I found that I had a number of $30^\circ$-$60^\circ$-$90^\circ$ triangles at my disposal. Working with various proportions among those would get to the result, but as I started to think that through, I realized that $R$ and $S$ were the ortho-/circum-/in-centers of equilateral triangles, and then everything fell into place nicely. So ... No, I didn't think about the equilaterals right away.
May 12, 2014 at 18:08 comment added rah4927 +1, brilliant solution.Thanks a lot.May I know the motivation behind the constructions?Was your main intention to create an equilateral triangle?
May 12, 2014 at 16:49 history edited Blue CC BY-SA 3.0
deleted 46 characters in body
May 12, 2014 at 16:39 history answered Blue CC BY-SA 3.0